On 15 April 2013 13:18, Thomas thomasfox...@aol.com wrote:
Dear List,
I am using both the clm() and clmm() functions from the R package 'ordinal'.
I am fitting an ordinal dependent variable with 5 categories to 9 continuous
predictors, all of which have been normalised (mean subtracted
Dear Sir,
Thanks a lot for your valuable input and guidance.
Regards
Katherine
--- On Mon, 15/4/13, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote:
From: Jeff Newmiller jdnew...@dcn.davis.ca.us
Subject: Re: [R] Sorting data.frame and again sorting within data.frame
To: David Winsemius
Hi
Do you mean ecdf? If yes just ose add option in plot.
plot(ecdf(rnorm(100, 1,2)))
plot(ecdf(rnorm(100, 2,2)), add=TRUE, col=2)
If not please specify from where is ecdf_stat or stat_ecdf which, as you
indicate, are the same functions.
Regrdas
Petr
-Original Message-
From:
Dear list,
I am trying to create a function from a string, and have so far solved
it with eval(parse()). This works well also when using the newly created
function as an argument to another function. The trouble starts when I
want to use it with parLapply. Below is a much simplified example:
Is this what you are looking for?
FUN = eval(bquote(function(x) .(parse(text = fstring)[[1]])))
FUN
function (x)
x + 2
FUN(3)
[1] 5
On Apr 16, 2013, at 09:50 , Jon Olav Skoien wrote:
Dear list,
I am trying to create a function from a string, and have so far solved it
with
Thanks a lot, that seems to do exactly what I need!
Best wishes,
Jon
On 16-Apr-13 10:21, peter dalgaard wrote:
Is this what you are looking for?
FUN = eval(bquote(function(x) .(parse(text = fstring)[[1]])))
FUN
function (x)
x + 2
FUN(3)
[1] 5
On Apr 16, 2013, at 09:50 , Jon Olav Skoien
I think it's your starting values for the initial state probability
distribution,
i.e. c(1,1,1)/3 that cause the problem. They seem to drop you into some
sort of local maximum/stationary point, a long way from the global maximum.
Try, e.g. c(4,2,1)/7; this gives me:
Hi Anonymous
There are different methods to select lags in unit roots tests, the two you
mention are not fundamentally wrong, and belong to the standard methods
used, even if the IC selection is maybe now the prefered solution. Note
there is some work from Perron and Ng with a refined selection
Dear Sir/Ma,
I Adelabu.A.A, one of the R-users from Nigeria. When am running a coxph command
the below error was generated, and have try some idea but not going through.
kindly please assist:
cox1 - coxph(Surv(tmonth,status) ~ sex + age + marital + sumassure, X)
Warning message:
In
It seems that indeed providing other starting values initiates iterations
to take place. However, more worrisome is that the does not seem to
converge, even when upping the number of iterations. Below I run a 2 state
model on the same as well for comparison (I have added set.seed statements
to
Looks like sumassure is treated as categorical. This sort of thing is usually a
data error; it happens if one of the values can not be converted to numeric, O
instead of 0, comma instead of period, etc.
Check summary(X), or, to investigate more specifically, things like
x - X$sumassure
Dear R forum
I have a data.frame
df = data.frame(currency_type = c(EURO_o_n, EURO_o_n, EURO_1w, EURO_1w,
USD_o_n, USD_o_n, USD_1w, USD_1w), rates = c(0.47, 0.475, 0.461, 0.464,
1.21, 1.19, 1.41, 1.43))
currency_type rates
1 EURO_o_n 0.470
2 EURO_o_n 0.475
3 EURO_1w
On Tue, Apr 16, 2013 at 8:38 AM, Katherine Gobin
katherine_go...@yahoo.com wrote:
Dear R forum
I have a data.frame
df = data.frame(currency_type = c(EURO_o_n, EURO_o_n, EURO_1w,
EURO_1w, USD_o_n, USD_o_n, USD_1w, USD_1w), rates = c(0.47, 0.475,
0.461, 0.464, 1.21, 1.19, 1.41, 1.43))
Hi,
Try:
df = data.frame(currency_type = c(EURO_o_n, EURO_o_n, EURO_1w, EURO_1w,
USD_o_n, USD_o_n, USD_1w, USD_1w), rates = c(0.47, 0.475, 0.461, 0.464,
1.21, 1.19, 1.41, 1.43),stringsAsFactors=FALSE)
df$currency-unlist(lapply(str_split(df[,1],_),`[`,1))
HI,
You can also do this by:
library(stringr)
df2-data.frame(currency=word(str_replace(df[,1],_, ),1),
temor=word(str_replace(df[,1],_, ),2),
rates=df$rates,stringsAsFactors=FALSE)
df2
# currency temor rates
#1 EURO o_n 0.470
#2 EURO o_n 0.475
#3 EURO 1w 0.461
#4 EURO
Hi,
I am not sure about the problem.
If your non-numeric vector is like:
a,b,,d,e,,f
vec1-unlist(str_split(readLines(textConnection(a,b,,d,e,,f)),,))
vec1[vec1==]- NA
vec1
#[1] a b NA d e NA f
If this doesn't work, please provide an example vector.
A.K.
Thanks for the response. That
Hi R users,
I have mentioned that R is getting slower if a process with a loop runs for
a while. Is that normal?
Let's say, I have a code which produce an output file after one loop run.
Now after 10, 15 or 20 loop runs the time between the created files is
stongly increasing.
Is there maybe any
On Apr 16, 2013, at 9:52 AM, Chris82 rubenba...@gmx.de wrote:
Hi R users,
I have mentioned that R is getting slower if a process with a loop runs for
a while. Is that normal?
Let's say, I have a code which produce an output file after one loop run.
Now after 10, 15 or 20 loop runs the
Hello,
The help page is pretty clear, I think. You have to pass an object of
class 'Arima', 'ar' or 'ets' to simulate.Arima.
See, for instance the second example in the help page for ?Arima. And
extend it like this:
set.seed(6816)
lines(simulate(air.model, nsim = 48), col = red)
Hope this
Dear Matthieu,
Many thanks for your reply.
I was not sure what the best way forward in selecting lag length. Eventually
I wrote a function that carries out serial correl test and AIC based lag
length selections. I used urca package.
Here is what I come up with in the end:
zamod.A=ur.za(x,
Hi all,
At RStudio, we're hosting our Introduction to R Workshop this May in
two locations. As an R-help subscriber, we're offering 10% off!
* Intro to data science with R (http://goo.gl/bplg3)
May 13-14 New York City
* Intro to data science with R (http://goo.gl/VCUFL)
May 20-21 San
#this is my data set.
data_set-data.frame(x0=c(1,1,0,0),
x1=c(1,1,0,0),x2=c(1,1,0,0),x3=c(1,1,0,0),x4=c(1,1,0,0))
#this is my target
target-c(1,1,0,0)
rf.vs1 - varSelRF(data_set, as.factor(target), ntree = 500, ntreeIterat =
300,
vars.drop.frac = 0.2)
rf.vs1
rf.vs1[[3]]
It
Hello,
I'm new to R and to Spatial Analysis and got a problem trying to create a
Spatial Weights Matrix.
*I us the following code to create the Neighbourslist:*
library(maptools)
library(spdep)
library(rgdal)
location_County- readShapePoly()
proj4string(location_County)- CRS(+proj=longlat
Hallo
Could somebody perhaps assist with my dilemma,
Package: VIF. The examples are not very clear (data is stored internally).
I wish to read a .csv file (header=TRUE) and run VIF. But I get
nonsensical output.
I have downloaded the boston.csv file (from the referring website).
How do I
Hi Arun,
This is excellent and elegant. I thought there had to be a relatively simple
way to do this. Thank you very much.
Jeremy
From: arun kirshna [via R] [mailto:ml-node+s789695n4664328...@n4.nabble.com]
Sent: Monday, April 15, 2013 10:34 PM
To: Crowley, Jeremy
Subject: Re: matching
Hadley:
I don't think this is appropriate. Think of what it would be like if everyone
shilled their R training and consulting wares here.
Bert
Sent from my iPhone -- please excuse typos.
On Apr 16, 2013, at 8:09 AM, Hadley Wickham h.wick...@gmail.com wrote:
Hi all,
At RStudio, we're
On Apr 16, 2013, at 12:45 AM, PIKAL Petr wrote:
Hi
Do you mean ecdf? If yes just ose add option in plot.
plot(ecdf(rnorm(100, 1,2)))
plot(ecdf(rnorm(100, 2,2)), add=TRUE, col=2)
If not please specify from where is ecdf_stat or stat_ecdf which, as you
indicate, are the same
Hi Milan and Max,
Thanks to each of you for your reply to my post. Thus far, I've managed to find
answers to some of the questions I asked initially.
I am now able to control the justification of the leftmost column in my tables,
as well as to add borders to the top and bottom. I also
Hi,
I have some data, that when plotted looks very close to a log-normal
distribution. My goal is to build a regression model to test how this variable
responds to several independent variables.
To do this, I want to use the fitdistr tool from the MASS package to see how
well my data fits
Given that we occasionally run into problems with comparing Excel
results to R results, and other spreadsheet-induced errors, I thought
this might be of interest.
http://www.nextnewdeal.net/rortybomb/researchers-finally-replicated-reinhart-rogoff-and-there-are-serious-problems
The punchline:
If
On 16/04/2013 1:19 PM, Noah Silverman wrote:
Hi,
I have some data, that when plotted looks very close to a log-normal
distribution. My goal is to build a regression model to test how this variable
responds to several independent variables.
To do this, I want to use the fitdistr tool from
Dear all,
What is the quickest and most efficient way to diff two data frames,
so as to obtain a vector of indices (or logical) for rows/columns that
differ in the two data frames? For example,
Xe - head(mtcars)
Xf - head(mtcars)
Xf[2:4,3:5] - 55
all.equal(Xe, Xf)
[1] Component 3: Mean
Hi,
Please check your dput().
By using your dput() output, I am getting:
$patient_id
[1] 2 2 2 2 3 3 3 3
$responsed_at
[1] 14755 14797 14835 14883 14755 14789 14826 14857
$number
[1] 1 2 3 4 1 2 3 4
$score
[1] 1 1 2 3 1 5 4 5
$.Names
[1] patient_id responsed_at number scores
Hi All,
Apologies if this has been answered somewhere else, but I have been
searching for an answer all day and not been able to find one.
I am trying to plot a path diagram for a CFA I have run, I have installed
Rgraphviz and run the following:
pathDiagram(cfa, min.rank='item1, item2, item3,
I thought I've understood the 'order' function, using simple examples like:
order(c(5,4,-2))
[1] 3 2 1
However, I arrived to the following example:
order(c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505, 1045))
[1] 8 9 10 7 11 6 5 4 3 2 1
and I was completely
On Apr 16, 2013, at 11:44 AM, Trina Patel trinarpa...@gmail.com wrote:
Hi,
I installed R 3.0.0 on a Windows 2008 Server.
When I submitted the following code in R64,
library(tools)
testInstalledBasic(scope=devel)
I get the following message in the R Console:
library(tools)
Hi Julio,
On Tue, Apr 16, 2013 at 1:51 PM, Julio Sergio julioser...@gmail.com wrote:
I thought I've understood the 'order' function, using simple examples like:
order(c(5,4,-2))
[1] 3 2 1
However, I arrived to the following example:
order(c(2465, 2255, 2085, 1545, 1335, 1210,
Hello,
Inline.
Em 16-04-2013 18:51, Julio Sergio escreveu:
I thought I've understood the 'order' function, using simple examples like:
order(c(5,4,-2))
[1] 3 2 1
However, I arrived to the following example:
order(c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505, 1045))
Hi,
vec1- c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505, 1045)
vec1[order(vec1)]
#[1] 210 210 505 920 1045 1210 1335 1545 2085 2255 2465
order(vec1)
#[1] 8 9 10 7 11 6 5 4 3 2 1
sort(vec1,index.return=TRUE)
#$x
#[1] 210 210 505 920 1045 1210 1335 1545 2085 2255 2465
On 16/04/2013 1:51 PM, Julio Sergio wrote:
I thought I've understood the 'order' function, using simple examples like:
order(c(5,4,-2))
[1] 3 2 1
However, I arrived to the following example:
order(c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505, 1045))
[1] 8 9 10 7
When in doubt, assume the spreadsheet is wrong. I suggested this to someone
have a problem with R vs Excel results a while ago. When I checked back with
him -- there was a spreadsheet error.
I think a t-shirt with the motto Friends don't let friends use
spreadsheets[1] sounds like a good
-Original Message-
From: gunter.ber...@gene.com
Sent: Tue, 16 Apr 2013 09:43:14 -0700
To: h.wick...@gmail.com
Subject: Re: [R] 10% off Intro R training from RStudio: NYC May 13-14, SF
May 20-21
Hadley:
I don't think this is appropriate. Think of what it would be like if
Julio Sergio juliosergio at gmail.com writes:
I thought I've understood the 'order' function, using simple examples like:
Thanks to you all!... As Sarah said, what was damaged was my understanding (
;-) )... and as Duncan said, I was confusing 'order' with 'rank',
thanks! Now I understand
On Tue, Apr 16, 2013 at 5:43 PM, Bert Gunter gunter.ber...@gene.com wrote:
Hadley:
I don't think this is appropriate. Think of what it would be like if everyone
shilled their R training and consulting wares here.
Everyone does, don't they? A search on Nabble shows up regular
postings from
[See in-line below[
On 16-Apr-2013 17:51:41 Julio Sergio wrote:
I thought I've understood the 'order' function, using simple examples like:
order(c(5,4,-2))
[1] 3 2 1
However, I arrived to the following example:
order(c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505,
Hello,
Maybe Petr Savicky's answer in the link
https://stat.ethz.ch/pipermail/r-help/2012-February/304830.html
can lead you to what you want.
I've changed his function a bit in order to return a logical vector
into the rows where different rows return TRUE.
setdiffDF2 - function(A, B){
Hi Bert: given what Hadley and Rstudio have provided to the R-community,
what's the big deal of
letting people know about a class. It's the ideal place to send the notice.
and yes, as Barry
and John said, every other commercial entity does send to the R-list.
Mark
On Tue, Apr 16, 2013 at
What a terrific article. Thanks for sharing! The more we critically
examine how research is actually done the more frightened we become.
Frank
--
Frank E Harrell Jr Professor and Chairman School of Medicine
Department of Biostatistics Vanderbilt University
Hi Bert,
We are following the mailing list guidelines to the best of our
knowledge (e.g.
http://r.789695.n4.nabble.com/R-development-master-class-NYC-Dec-12-13-td4037031.html#a4038699).
It's our belief (as shared by others) that advertising our courses
falls under the general aegis of helping
Dear all,
How should I add several variables to a data frame without losing the
attributes of the df? Consider the following:
require(Hmisc)
Xa - iris
label(Xa, self=T) - Some df label
str(Xa)
'data.frame': 150 obs. of 5 variables:
$ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9
I tend to live in fear that some spreadsheet calculating a drug dose for me
will use my telephone number rather than my weight.
John Kane
Kingston ON Canada
-Original Message-
From: f.harr...@vanderbilt.edu
Sent: Tue, 16 Apr 2013 13:20:46 -0500
To: r-h...@stat.math.ethz.ch
Hi,
I'm actually trying to rank a set of candidate models with an information
criterion (AICc, QIC, BIC). The problem I have is that I use mixed-effect cox
regression only available with the package {coxme} (see the example below).
#Model1
spring.cox - coxme (Surv(start, stop, Real_rand) ~
I think Duncan said that order and rank were inverses (if there are no ties).
order() has
period 2 so order(order(x)) is also rank(x) if there are no ties. E.g.,
data.frame(x, o1=order(x), o2=order(order(x)), o3=order(order(order(x))),
o4=order(order(order(order(x, rank=rank(x))
x
Of course but you should carefully read the guidelines (see bottom of post and
it is a good idea to read
Reproducibility
https://github.com/hadley/devtools/wiki/Reproducibility
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
for some useful suggestions on
HI,
Not sure if this helps:
library(plyr)
res-mutate(Xa,var1=round(Sepal.Length),var2=round(Sepal.Width))
str(res)
#'data.frame': 150 obs. of 7 variables:
# $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
# $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
# $
Dear Laura,
This works for me. Is dot on your system path?
Best,
John
---
John Fox
Senator McMaster Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
-Original Message-
From:
Are my emails getting through?
2013/4/14 santiago gil sg.c...@gmail.com:
Hello all,
I have a problem with the way attributes are dealt with in the
function xmlToList(), and I haven't been able to figure it out for
days now.
Say I have a document (produced by nmap) like this:
mydoc -
On Apr 16, 2013, at 12:43, Bert Gunter gunter.ber...@gene.com wrote:
Hadley:
I don't think this is appropriate. Think of what it would be like if everyone
shilled their R training and consulting wares here.
Echoing others, this seems an accepted practice on the lists, endorsed at least
Yes. This is the third such copy. You can view them all in the Archive,
starting with the first one:
https://stat.ethz.ch/pipermail/r-help/2013-April/351504.html
On Apr 16, 2013, at 11:49 AM, santiago gil wrote:
Are my emails getting through?
2013/4/14 santiago gil sg.c...@gmail.com:
Hi, Santiago:
Yes, your e-mail has been received. I'm sorry, I can't solve your question.
Regards.
Eva
--- El mar, 16/4/13, santiago gil sg.c...@gmail.com escribió:
De: santiago gil sg.c...@gmail.com
Asunto: Re: [R] Problem with handling of attributes in xmlToList in XML package
Para:
On Wed, Apr 17, 2013 at 5:19 AM, Noah Silverman noahsilver...@ucla.eduwrote:
Hi,
I have some data, that when plotted looks very close to a log-normal
distribution. My goal is to build a regression model to test how this
variable responds to several independent variables.
[snip]
When I
On Apr 16, 2013, at 6:38 AM, arun wrote:
Hi,
I am not sure about the problem.
If your non-numeric vector is like:
a,b,,d,e,,f
vec1-unlist(str_split(readLines(textConnection(a,b,,d,e,,f)),,))
vec1[vec1==]- NA
vec1
#[1] a b NA d e NA f
If this doesn't work, please provide an
Quantreggers:
I'm trying to run rq() on a dataset I posted at:
https://docs.google.com/file/d/0B8Kij67bij_ASUpfcmJ4LTFEUUk/edit?usp=sharing
(it's a 1500kb csv file named singular.csv) and am getting the following
error:
mydata - read.csv(singular.csv)
fit_spl - rq(raw_data[,1] ~
Hi,
On Apr 16, 2013, at 2:49 PM, santiago gil wrote:
2013/4/14 santiago gil sg.c...@gmail.com:
Hello all,
I have a problem with the way attributes are dealt with in the
function xmlToList(), and I haven't been able to figure it out for
days now.
I have not used xmlToList(), but I find
@Duncan, You make a very good point. Somehow I overlooked that 0 is not
positive. I guess that rules out the log normal model.
My challenge here is finding the right model for this data. Originally it was
a nice count of students. Relatively easy to model with a zero inflated
Poisson
Le 16/04/13 15:52, Chris82 a écrit :
Hi R users,
I have mentioned that R is getting slower if a process with a loop runs for
a while. Is that normal?
Let's say, I have a code which produce an output file after one loop run.
Now after 10, 15 or 20 loop runs the time between the created files is
Noah,
You might want to look at beta regression, using the betareg package on CRAN.
There is a JSS paper here that you might find helpful:
http://www.jstatsoft.org/v34/i02/paper
along with the vignettes for the package:
http://cran.r-project.org/web/packages/betareg/vignettes/betareg.pdf
Hi,
Another method would be:
Xc- Xa
Xc$var1-NA; Xc$var2- NA
Xc[]- append(as.list(Xa),as.list(Xb))
str(Xc)
#'data.frame': 150 obs. of 7 variables:
# $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
# $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
# $ Petal.Length:
Just to add:
Xc[]- append(Xa,Xb) #should also work
str(Xc)
#'data.frame': 150 obs. of 7 variables:
# $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
# $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
# $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
On Apr 16, 2013, at 22:20 , Noah Silverman wrote:
@Duncan, You make a very good point. Somehow I overlooked that 0 is not
positive. I guess that rules out the log normal model.
My challenge here is finding the right model for this data. Originally it
was a nice count of students.
Hi Marc,
Thank you for the links to all the resources, I will be sure to review
them in detail.
As for running,
Sys.setenv(LC_COLLATE = C, LANGUAGE = en)
I'm sorry that I forgot to mention that I did set the above
enviornmental variables as specified. Both within R, as suggested in
your email,
On 04/17/2013 03:25 AM, Sarah Goslee wrote:
...
Ouch.
(Note: I know nothing about the site, the author of the article, or
the study in question. I was pointed to it by someone else. But if
true: highly problematic.)
Sarah
There seem to be three major problems described here, and only one is
Dear List,
I've just tried to specify a GAM without an intercept -- I've got one
of the (rare) cases where it is appropriate for E(y) - 0 as X -0.
Naively running a GAM with the -1 appended to the formula and the
calling predict.gam, I see that the model isn't behaving as expected.
I
Dear List,
I've just tried to specify a GAM without an intercept -- I've got one of
the (rare) cases where it is appropriate for E(y) - 0 as X -0.
Naively running a GAM with the -1 appended to the formula and the
calling predict.gam, I see that the model isn't behaving as expected.
I don't
William Dunlap wdunlap at tibco.com writes:
I think Duncan said that order and rank were inverses (if there are no
ties). order() has
period 2 so order(order(x)) is also rank(x) if there are no ties. E.g.,
Thanks William! This is very interesting. So, applying order two times I can
please deleter this thread -- wrong title
On 04/16/2013 02:35 PM, Andrew Crane-Droesch wrote:
Dear List,
I've just tried to specify a GAM without an intercept -- I've got one
of the (rare) cases where it is appropriate for E(y) - 0 as X -0.
Naively running a GAM with the -1 appended to the
Have you looked at the result of
bs(raw_data[,i], df=15)
? If there are not many unique values in the input there
will be a lot of NaN's in the output (because there are
repeated knots) and those NaN's will cause rq() to give
that message.
E.g.,
d - data.frame(y=sin(1:100),
Hi,
I want to plot two variables on the same graph but with two y axis just
like what you can do in Excel. I searched online that seems like you can
not achieve that in ggplot. So is there anyway I can do it in a nice way in
basic plot?
Suppose my data looks like this:
WeightHeight Date
I apologize for the multiple posting then, it's just that I received
those emails saying that my post was awaiting approval and more than
four days went by without news. Sorry for the lack of patience.
Thank you very much, Ben. Indeed that's how I've been doing it so far,
but I have accrued too
On 04/17/2013 08:35 AM, Ye Lin wrote:
Hi,
I want to plot two variables on the same graph but with two y axis just
like what you can do in Excel. I searched online that seems like you can
not achieve that in ggplot. So is there anyway I can do it in a nice way in
basic plot?
Suppose my data
peter dalgaard pdalgd at gmail.com writes:
On Apr 16, 2013, at 22:20 , Noah Silverman wrote:
My challenge here is finding the right model for this data.
Originally it was a nice count of students. Relatively easy to
model with a zero inflated Poisson model. The resulting residuals
Hello All,
Would anyone be able to help me understand how R computes a
quantile-quantile plot for comparing two data samples with unequal sample
sizes? Normally, the procedure should be to rearrange the larger data
sample into n equally-spaced parts using interpolation, where n is the
sample
Hello,
This is Elaine.
I am using R 3.0 to download package vegan but failed.
The warning message is
package vegan successfully unpacked and MD5 sums checked
Warning: unable to move temporary installation
C:\Users\elaine\Documents\R\win-library\3.0\file16c82da53b1b\vegan to
On Apr 16, 2013, at 20:12, Janh Anni annij...@gmail.com wrote:
Hello All,
Would anyone be able to help me understand how R computes a
quantile-quantile plot for comparing two data samples with unequal sample
sizes? Normally, the procedure should be to rearrange the larger data
sample
Hello All,
I manually moved the vegan.zip to C:\Users\elaine\Documents\R\
win-library\3.0\vegan.
Then unzipping the file.
It worked to require vegan
Elaine
On Wed, Apr 17, 2013 at 8:56 AM, Elaine Kuo elaine.kuo...@gmail.com wrote:
Hello,
This is Elaine.
I am using R 3.0 to download
Hi,
On Apr 16, 2013, at 6:39 PM, santiago gil wrote:
Thank you very much, Ben. Indeed that's how I've been doing it so far,
but I have accrued too many reasons not to work with the XML object
any more and move all my coding to a list formulation.
I wonder what you mean with
[...] but I
Hello Michael,
Thanks for that information.
Regards
Janh
On Tue, Apr 16, 2013 at 9:13 PM, Michael Weylandt
michael.weyla...@gmail.com wrote:
On Apr 16, 2013, at 20:12, Janh Anni annij...@gmail.com wrote:
Hello All,
Would anyone be able to help me understand how R computes a
HI Philippos,
Try this:
dat1- read.csv(Validation_data_set3.csv,sep=,,stringsAsFactors=FALSE)
#converted to csv
str(dat1)
#'data.frame': 12573 obs. of 17 variables:
# $ Removed.AGC : num 65.67 46.17 41.26 14.09
5.38 ...
# $ Removed.SST
Verstuurd vanaf mijn iPad
Bert Verleysen
00 32 (0)477 874 272
www.beverconsult.be
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
Hi,
I want to save a plot in the windows device as png and the default
resolution is 72dpi. Is it possible to increase the default resolution
to for example 300 dpi?
I have thought of using function png(..., res=300), but the problem is
that the figure produced this way looks different than
maybe something wrong with your R_LIBS(it should be R_LIBS=dir/R-
3.0.0/lib64/)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide
I have been using the following so far without having any problems:
dev.copy(png,sample.png,width=8, height=10, units=in,res=500)
dev.off()
On Tue, Apr 16, 2013 at 6:32 PM, jt...@mappi.helsinki.fi wrote:
Hi,
I want to save a plot in the windows device as png and the default
resolution is
Hi,
Do it yourself:
https://stat.ethz.ch/mailman/listinfo/r-help
Hint:
Bbottom of the page (To unsubscribe from R-help)
Regards,
Pascal
On 04/17/2013 06:33 AM, Bert Verleysen (beverconsult) wrote:
Verstuurd vanaf mijn iPad
Bert Verleysen
00 32 (0)477 874 272
www.beverconsult.be
Hi Farnoosh,
YOu can use either ?merge() or ?join()
DataA- read.table(text=
ID v1
1 10
2 1
3 22
4 15
5 3
6 6
7 8
,sep=,header=TRUE)
DataB- read.table(text=
ID v2
2 yes
5 no
7 yes
,sep=,header=TRUE,stringsAsFactors=FALSE)
HI,
I have a dataframe with two variable A, B. I transform the two variable
and name them as C, D and save it in a dataframe dfcd. However, I wonder
why can't I call them by dfcd$C and dfcd$D?
Thanks,
Miao
A=c(1,2,3)
B=c(4,6,7)
dfab-data.frame(A,B)
C=dfab[A]*2
D=dfab[B]*3
Hi,
Because a column name exists for C and D:
colnames(C)
[1] A
colnames(D)
[1] B
One possibility:
A=c(1,2,3)
B=c(4,6,7)
dfab-data.frame(A,B)
C=dfab$A*2
D=dfab$B*3
dfcd-data.frame(C,D)
dfcd
C D
1 2 12
2 4 18
3 6 21
dfcd$C
[1] 2 4 6
HTH,
Pascal
On 04/17/2013 02:33 PM, jpm miao
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